NCERT Solutions for Class 9 Maths Chapter 10 Circles

 

NCERT solutions for class 9 maths chapter 10 Circles:  A round figure made up using several points that are equidistant from a reference point is called a circle. You can relate this chapter with real-life things. Many objects that we come across in our daily life is circular in shape, such as ring, bangle, wheels of the vehicle, clock, etc. CBSE NCERT solutions for class 9 maths chapter 10 Circles will help in solving all problems related to circular shapes. The circle is an integral part of unit geometry. A circle divides the plane in which the circle lies into three parts as shown in figure 1, which are the interior of the circle, the exterior of the circle and the circle. In this particular chapter, you will learn the proof of theorems and problems based on those theorems. Solutions of NCERT for class 9 maths chapter 10 Circles are designed in such a way that a student can fetch 100% marks in a particular question. In the upcoming class, you will get to know some advanced versions of circles but the advanced part can only be done when you have done the basics. So, class 9 maths circles are a foundation for next class board examinations. NCERT solutions for class 9 maths chapter 10 circles will assist you in building this strong foundation. NCERT solutions are also available for different chapters, classes, and subjects which you can get by clicking on the given link. Six exercises of this chapter are explained below.

Exercise:10.1

Exercise:10.2

Exercise:10.3

Exercise:10.4

Exercise:10.5

Exercise:10.6

 
 

NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.1

Fill in the blanks:                                                                                                            

Q1 (i) The centre of a circle lies in _____________ of the circle. (exterior/ interior)

Answer:

The centre of a circle lies in the interior of the circle.

Fill in the blanks:    

Q1 (ii) A point, whose distance from the centre of a circle is greater than its radius lies in_____________ of the circle. (exterior/ interior)

Answer:

A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.

Fill in the blanks:

Q1 (iii) The longest chord of a circle is a ___________ of the circle.

Answer:

The longest chord of a circle is a  diameter of the circle.

Fill in the blanks:

Q1 (iv) An arc is a ________________ when its ends are the ends of a diameter.

Answer:

An arc is a semi- circle when its ends are the ends of a diameter.

Fill in the blanks:

Q1 (v) Segment of a circle is the region between an arc and ________________ of the circle.

Answer:

Segment of a circle is the region between an arc and chord of the circle.

Fill in the blanks:    

Q1 (vi) A circle divides the plane, on which it lies, in _______________ parts.

Answer:

A circle divides the plane, on which it lies, in two parts.

Write True or False: Give reasons for your answers.

Q2 (i) Line segment joining the centre to any point on the circle is a radius of the circle.

Answer:

True. As line segment joining the centre to any point on the circle is a radius of the circle.

Write True or False: Give reasons for your answers.

Q2 (ii) A circle has only finite number of equal chords.

Answer:

False . As a circle has infinite number of equal chords.

Write True or False: Give reasons for your answers.

Q2 (iii) If a circle is divided into three equal arcs, each is a major arc.

Answer:

False.  If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas major arc makes angle greater than 180 degree at centre.

Write True or False: Give reasons for your answers.

Q2 (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Answer:

True.A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Write True or False: Give reasons for your answers.

Q2 (v) Sector is the region between the chord and its corresponding arc.

Answer:

False. As the sector is the region between the radii and arc.

Write True or False: Give reasons for your answers.

Q2 (vi) A circle is a plane figure.

Answer:

True. A circle is a plane figure.

 

NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.2

Q1 Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer:

Given: The two circles are congruent if they have the same radii.

To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. \angleBAC=\angleQPR

Proof : 

In \triangleABC and \trianglePQR,

     BC = QR         (Given)

    AB = PQ          (Radii of congruent circle)

    AC = PR          (Radii of congruent circle)

Thus,  \triangleABC \cong \trianglePQR         (By SSS rule)

     \angleBAC=\angleQPR        (CPCT)

Q2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer:

Given : chords of congruent circles subtend equal angles at their centres,

To prove :  BC = QR

Proof : 

In \triangleABC and \trianglePQR,

     \angleBAC=\angleQPR       (Given)

    AB = PQ          (Radii of congruent circle)

    AC = PR          (Radii of congruent circle)

Thus,  \triangleABC \cong \trianglePQR         (By SAS rule)

      BC = QR             (CPCT)

 

NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.3

Q1 Draw different pairs of circles. How many points does each pair have in common? What ii the maximum number of common points?

Answer:

In (i) we do not have any common point.

In (ii) we have 1 common point.

In (iii) we have 1 common point.

In (iv) we have 2 common points.

The maximum number of common points is 2.

Q2 Suppose you are given a circle. Give a construction to find its centre.
Answer:

 

Given : Points P,Q,R lies on circle.

Construction :

                 1. Join PR and QR

                 2. Draw perpendicular bisector of PR and QR which intersects at point O.

                 3. Taking O as centre and OP as radius draw a circle.

                 4. The circle obtained is required.

Q3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer:

Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

             

Construction: Joinpoint P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r)  and QM is the bisector of chord AB.

    \therefore PM\perp AB

Hence, \angle PMA =90 \degree

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

   \therefore QM\perp AB

Hence, \angle QMA =90 \degree

Now, \angle QMA +\angle PMA=90 \degree+90 \degree= 180 \degree

\anglePMA and \angleQMA are forming linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

 

NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.4

Q1 Two circles of radii \small 5\hspace{1mm}cm and \small 3\hspace{1mm}cm intersect at two points and the distance between their  centres is \small 4\hspace{1mm}cm. Find the length of the common chord.
Answer:

Given: Two circles of radii \small 5\hspace{1mm}cm and \small 3\hspace{1mm}cm intersect at two points and the distance between their centres is \small 4\hspace{1mm}cm.

To find the length of the common chord.

Construction: Join OP and draw OM\perp AB\, \, and \, \, \, ON\perp CD.

Proof: AB is a chord of circle C(P,3)  and PM is the bisector of chord AB.

           \therefore PM\perp AB

             \angle PMA=90 \degree

Let, PM = x , so QM=4-x

In \triangleAPM, using Pythagoras theorem

          AM^2=AP^2-PM^2...........................1

Also, 

In \triangleAQM, using Pythagoras theorem

          AM^2=AQ^2-MQ^2...........................2

From 1 and 2, we get

        AP^2-PM^2=AQ^2-MQ^2

\Rightarrow 3^2-x^2=5^2-(4-x)^2

\Rightarrow 9-x^2=25-16-x^2+8x

\Rightarrow 9=9+8x

\Rightarrow 8x=0

\Rightarrow x=0

Put,x=0 in equation 1

          AM^2=3^2-0^2=9

       \Rightarrow AM=3

\Rightarrow AB=2AM=6

 

Q2 If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answer:

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP=DP.

Construction : Join OP and draw OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.

Proof : 

        

 

In \triangleOMP and \triangleONP,

     AP = AP         (Common)

      OM = ON          (Equal chords of a circle are equidistant from the centre)

      \angleOMP = \angleONP      (Both are right angled)

Thus,  \triangleOMP \cong \triangleONP         (By SAS rule)

      PM = PN..........................1 (CPCT)

      AB = CD ............................2(Given )

\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD

\Rightarrow AM = CN......................3

Adding 1 and 3, we have

 AM + PM = CN + PN

 \Rightarrow AP = CP

Subtract 4 from 2, we get

 AB-AP = CD - CP

\Rightarrow PB = PD

 

Q3 If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer:

Given: two equal chords of a circle intersect within the circle.

To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. \angleOPM=\angleOPN

Proof : 

 

Construction: Join OP and draw OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.

In \triangleOMP and \triangleONP,

        AP = AP         (Common)

      OM = ON          (Equal chords of a circle are equidistant from the centre)

      \angleOMP = \angleONP      (Both are right-angled)

Thus,  \triangleOMP \cong \triangleONP         (By RHS rule)

                 \angleOPM=\angleOPN   (CPCT)

Q4 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that \small AB=CD (see Fig. \small 10.25).
 

            

Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B,  C and D.

To prove : AB  = CD

Construction: Draw OM\perp AD

Proof : 

BC is a chord of the inner circle and OM\perp BC

So, BM = CM  .................1

(Perpendicular OM bisect BC)

Similarly,

AD  is a chord of the outer circle and OM\perp AD

So, AM = DM  .................2

(Perpendicular OM bisect AD )

Subtracting   1 from  2, we get

 AM-BM = DM - CM

\Rightarrow AB = CD

 

Q5 Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius   \small 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma.  If the distance between Reshma and Salma and between Salma and Mandip is \small 6m each, what is the distance between Reshma and Mandip?

Answer:

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

        So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw OL\perp RS.

Proof: 

             

In \triangle ORS,

OS = OR    and  OL\perp RS (by construction )

So,       RL = LS = 3cm        (RS = 6 cm )

In \triangleOLS, by pytagoras theorem,

OL^2=OS^2-SL^2

\Rightarrow OL^2=5^2-3^2=25-9=16

\Rightarrow OL=4

In   \triangle ORK and \triangleOMK,

   OR  = OM  (Radii)

   \angleROK = \angle MOK    (Equal chords subtend equal angle at centre)

         OK = OK         (Common)

     \triangle ORK \cong \triangleOMK    (By SAS)

       RK = MK    (CPCT)

Thus,OK\perp RM

area of \triangleORS  =\frac{1}{2}\times RS\times OL...............................1

area of \triangleORS  =\frac{1}{2}\times OS\times KR.............................2

From 1 and 2, we get

\frac{1}{2}\times RS\times OL  =\frac{1}{2}\times OS\times KR

\Rightarrow RS\times OL=OS\times KR

\Rightarrow 6\times 4=5\times KR

\Rightarrow KR=4.8 cm

Thus, RM =2 KR=2\times 4.8 cm=9.6 cm

 

Q6 A circular park of radius \small 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD 

Radius of circular park = 20 m

  so, AO=SO=DO=20 m

Construction: AP\perp SD

Proof :

       

Let AS = SD = AD = 2x cm

In \triangleASD,

  AS = AD   and     AP\perp SD

So,  SP = PD = x cm 

In \triangleOPD, by Pythagoras,

       OP^2=OD^2-PD^2

\Rightarrow OP^2=20^2-x^2=400-x^2

\Rightarrow OP=\sqrt{400-x^2}

In \triangleAPD, by Pythagoras,

   AP^2=AD^2-PD^2

\Rightarrow (AO+OP)^2+x^2=(2x)^2

\Rightarrow (20+\sqrt{400-x^2})^2+x^2=4x^2

\Rightarrow 400+400-x^2+40\sqrt{400-x^2}+x^2=4x^2

\Rightarrow 800+40\sqrt{400-x^2}=4x^2

\Rightarrow 200+10\sqrt{400-x^2}=x^2

\Rightarrow 10\sqrt{400-x^2}=x^2-200

Squaring both sides,

\Rightarrow 100(400-x^2)=(x^2-200)^2

\Rightarrow 40000-100x^2=x^4-40000-400x^2

\Rightarrow x^4-300x^2=0

\Rightarrow x^2(x^2-300)=0

\Rightarrow x^2=300

\Rightarrow x=10\sqrt{3}

Hence, length of string of each phone= 2x=20\sqrt{3}m

 

NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.5

Q1 In Fig.  \small 10.36, A,B and C are three points on a circle with centre  O such that \small \angle BOC=30^{\circ}  and  \small \angle AOB=60^{\circ}. If D is a point on the circle other than the arc ABC, find \small \angle ADC.

            

Answer:

\angleAOC = \angleAOB + \angleBOC= 60 \degree+30 \degree=90 \degree

\angleAOC = 2 \angleADC   (angle subtended by an arc at the centre is double the angle subtended by it at any)

\angle ADC=\frac{1}{2}\angle AOC

\Rightarrow \angle ADC=\frac{1}{2}90 \degree= 45 \degree

 

Q2 A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer:

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and \angleACB.

Solution :

             

In \triangleOAB,

               OA = AB        (Given )

              OA = OB          (Radii of circle)

So,  OA=OB=AB

\Rightarrow ABC is a equilateral triangle.

So,\angleAOB = 60 \degree

  \angleAOB  = 2 \angleADB 

\Rightarrow \angle ADB=\frac{1}{2}\angle AOB

\Rightarrow \angle ADB=\frac{1}{2}60 \degree=30

ACBD is a cyclic quadrilateral .

So, \angleACB+\angleADB =180 \degree

   \Rightarrow \angle ACB+30 \degree= 180 \degree

\Rightarrow \angle ACB= 180 \degree-30 \degree=150 \degree

 

Q3 In Fig. \small 10.37 ,  \small \angle PQR=100^{\circ}, where P, Q and R are points on a circle with centre O.  Find \small \angle OPR.

             

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, \anglePSR + \anglePQR = 180 \degree

\Rightarrow \angle PSR+100 \degree=180 \degree

\Rightarrow \angle PSR=180 \degree-100 \degree=80 \degree

Here, \anglePOR = 2 \anglePSR

\Rightarrow \angle POR=2\times 80 \degree=160 \degree

In \triangleOPR , 

          OP=OR    (Radii )

          \angleORP = \angleOPR    (the angles opposite to equal sides)

In \triangleOPR , 

       \angleOPR+\angleORP+\anglePOR=180 \degree

\Rightarrow 2\angle OPR+160 \degree= 180 \degree

\Rightarrow 2\angle OPR=180 \degree- 160 \degree

\Rightarrow 2\angle OPR=20 \degree

\Rightarrow \angle OPR=10 \degree

 

Q4 In Fig. \small 10.38, \small \angle ABC=69^{\circ}, \angle ACB=31^{\circ}, find \small \angle BDC


           

Answer:

In \triangle ABC,

        \angleA+\angleABC+\angleACB=180\degree

\Rightarrow \angle A+69 \degree+31 \degree=180\degree

\Rightarrow \angle A+100 \degree=180\degree

\Rightarrow \angle A=180 \degree-100\degree

\Rightarrow \angle A=80 \degree

\angleA  = \angleBDC = 80 \degree   (Angles in same segment)

Q5 In Fig. \small 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that  \small \angle BEC=130^{\circ}  and \small \angle ECD=20^{\circ}. Find \small \angle BAC

        .

Answer:

\angleDEC+ \angleBEC = 180 \degree          (linear pairs)

\Rightarrow \angleDEC+ 130 \degree = 180 \degree           (\angle BEC =130 \degree)

\Rightarrow \angleDEC = 180 \degree - 130 \degree

\Rightarrow \angleDEC = 50 \degree

In \triangle DEC,

\angleD+ \angleDEC+ \angleDCE = 180 \degree

\Rightarrow \angle D+50 \degree+20 \degree= 180 \degree

\Rightarrow \angle D+70 \degree= 180 \degree

\Rightarrow \angle D= 180 \degree-70 \degree=110 \degree

  

\angleD = \angleBAC     (angles in same segment are equal  )

 \angleBAC  = 110 \degree

 

Q6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If  \small \angle DBC=70^{\circ}\small \angle BAC is \small 30^{\circ}, find \small \angle BCD. Further, if \small AB=BC, find \small \angle ECD.

Answer:

\angle BDC=\angle BAC       (angles in the same segment are equal )

\angle BDC= 30 \degree

In \triangle BDC,

       \angle BCD+\angle BDC+\angle DBC= 180 \degree

\Rightarrow \angle BCD+30 \degree+70 \degree= 180 \degree

\Rightarrow \angle BCD+100 \degree= 180 \degree

\Rightarrow \angle BCD=180 \degree- 100 \degree=80 \degree

If  AB = BC ,then 

    \angle BCA=\angle BAC

\Rightarrow \angle BCA=30 \degree

Here, \angle ECD+\angle BCE=\angle BCD

   \Rightarrow \angle ECD+30 \degree=80 \degree

\Rightarrow \angle ECD=80 \degree-30 \degree=50 \degree

 

Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer:

AC is the diameter of the circle.

Thus,\angle ADC=90 \degree   and    \angle ABC=90 \degree............................1(Angle in a semi-circle is right angle)

Similarly, BD is the diameter of the circle.

 Thus,\angle BAD=90 \degree   and    \angle BCD=90 \degree............................2(Angle in a semi-circle is right angle)

From 1 and 2, we get

          \angle BCD=\angle ADC=\angle ABC=\angle BAD =90 \degree

Hence, ABCD is a rectangle.

Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

                      AB || DE    (Given )

                      AD || BE    ( By construction )

Thus, ABED is a parallelogram.

     AD = BE     (Opposite sides of parallelogram )

    AD = BC      (Given )

so,  BE = BC

In \triangleEBC,

           BE = BC    (Proved above )

      Thus, \angle C = \angle 2...........1(angles opposite to equal sides )

              \angle A= \angle 1...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

    \angle 1+\angle 2=180 \degree           (linear pair)

     \Rightarrow \angle A+\angle C=180 \degree

Thus, ABED is a cyclic quadrilateral.

 

Q9 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. \small 10.40). Prove that \small \angle ACP=\angle QCD.

     

 

Answer:

\angle ABP=\angle QBD................1(vertically opposite angles)

\angle ACP=\angle ABP..................2(Angles in the same segment are equal)

\angle QBD=\angle QCD.................3(angles in the same segment are equal)

From 1,2,3 ,we get

\angle ACP=\angle QCD

Q10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: AB is the diameter of the circle and \angleADB is formed in a semi-circle.

             \angleADB = 90 \degree........................1(angle in a semi-circle)

Similarly,          

AC is the diameter of the circle and \angleADC is formed in a semi-circle.

             \angleADC = 90 \degree........................2(angle in a semi-circle)

From 1 and 2, we have

      \angleADB+\angleADC=90 \degree+90 \degree=180 \degree

\angleADB and \angleADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

 

Q11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
\small \angle CAD =\angle CBD.

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : \small \angle CAD =\angle CBD

Proof :

             

Triangle ABC and ADC are on common base BC and \angleBAC = \angleBDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

     \angleCAD = \angleCBD  (Angles in same segment are equal)

 

Q12 Prove that a cyclic parallelogram is a rectangle.
Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

            

  In cyclic quadrilateral ABCD.

            \angle A + \angle C = 180 \degree.......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

           \angle A = \angle C........................................2(opposite angles of a parallelogram are equal )

        From 1 and 2,

        \angle A + \angle A = 180 \degree

  \Rightarrow 2\angle A = 180 \degree

 \Rightarrow \angle A = 90 \degree

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

 

NCERT solutions for class 9 maths chapter 10 Circles Excercise: 10.6

Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer:

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : \anglePAQ = \anglePBQ 

Proof : In \triangleAPQ and \triangleBPQ,

                  PA = PB         (radii of same circle)

                 PQ = PQ        (Common)

                  QA = QB       (radii of same circle)

     So,           \triangleAPQ \cong \triangleBPQ       (By SSS)

                 \anglePAQ = \anglePBQ     (CPCT)

 

Q2 Two chords AB and CD of lengths \small 5\hspace {1mm}cm and \small 11\hspace {1mm}cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is \small 6\hspace {1mm}cm, find the radius of the circle.

Answer:

Given : AB = 5 cm, CD = 11 cm and AB || CD.

To find Radius (OA).

Construction: Draw OM \perp CD \, \, and \, \, \, ON\perp AB

Proof : 

          

Proof: CD is a chord of circle  and  OM \perp CD

    Thus, CM = MD = 5.5 cm    (perpendicular from centre bisects chord)

 and      AN = NB  = 2.5 cm

       Let OM be x.

 So, ON = 6 - x              (MN = 6 cm )

In \triangle OCM , using Pythagoras,

                 OC ^2=CM^2+OM^2.............................1

and

In \triangle OAN , using Pythagoras,

                 OA ^2=AN^2+ON^2.............................2

From 1 and 2,

      CM ^2+OM^2=AN^2+ON^2            (OC=OA =radii)

      5.5 ^2+x^2=2.5^2+(6-x)^2

\Rightarrow 30.25+x^2=6.25+36+x^2-12x

\Rightarrow 30.25-42.25=-12x

\Rightarrow -12=-12x

\Rightarrow x=1

From 2, we get 

OC^2=5.5^2+1^2=30.25+1=31.25

\Rightarrow OC=\frac{5}{2}\sqrt{5} cm

OA = OC 

Thus, the radius of the circle is \frac{5}{2}\sqrt{5} cm

Q3 The lengths of two parallel chords of a circle are \small 6\hspace {1mm}cm and \small 8\hspace {1mm}cm. If the smaller chord is at distance \small 4\hspace {1mm}cm from the centre, what is the distance of the other chord from the centre?

Answer:

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw OM \perp CD \, \, and \, \, \, ON\perp AB

Proof : 

          

Proof: CD is a chord of circle  and  OM \perp CD

    Thus, CM = MD = 3 cm    (perpendicular from centre bisects chord)

 and      AN = NB  = 4 cm

       Let MN be x.

 So, ON = 4 - x              (MN = 4 cm )

In \triangle OCM , using Pythagoras,

                 OC ^2=CM^2+OM^2.............................1

and

In \triangle OAN , using Pythagoras,

                 OA ^2=AN^2+ON^2.............................2

From 1 and 2,

      CM ^2+OM^2=AN^2+ON^2            (OC=OA =radii)

      \Rightarrow 3 ^2+4^2=4^2+(4-x)^2

\Rightarrow 9+16=16+16+x^2-8x

\Rightarrow 9=16+x^2-8x

\Rightarrow x^2-8x+7=0

\Rightarrow x^2-7x-x+7=0

\Rightarrow x(x-7)-1(x-7)=0

\Rightarrow (x-1)(x-7)=0

\Rightarrow x=1,7

 So, x=1  (since  x\neq 7> OM)

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that \small \angle ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove : \angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)

Construction: Join AC and DE.

Proof : 

            

Let \angleADC = x , \angleDOE = y and  \angleAOD = z

So,  \angleEOC = z (each chord subtends equal angle at centre)

   \angleAOC + \angleDOE +\angleAOD + \angleEOC = 360 \degree

\Rightarrow x+y+z+z=360 \degree

\Rightarrow x+y+2z=360 \degree.........................................1

In \triangle OAD ,

        OA = OD  (Radii of the circle)

       \angleOAD = \angleODA    (angles opposite to equal sides )

    \angleOAD + \angleODA + \angleAOD =180 \degree

\Rightarrow 2\angle OAD+z=180 \degree

\Rightarrow 2\angle OAD=180 \degree-z

\Rightarrow \angle OAD=\frac{180 \degree-z}{2}

\Rightarrow \angle OAD=90 \degree-\frac{z}{2}.............................................................2

Similarly,

\Rightarrow \angle OCE=90 \degree-\frac{x}{2}.............................................................3

\Rightarrow \angle OED=90 \degree-\frac{y}{2}..............................................................4

\angleODB is exterior of triangle OAD . So,

 \angleODB = \angleOAD + \angleODA

\Rightarrow \angle ODB=90 \degree-\frac{z}{2}+z                  (from  2)

\Rightarrow \angle ODB=90 \degree+\frac{z}{2}.................................................................5

similarly,

\angleOBE is exterior of triangle OCE . So,

 \angleOBE = \angleOCE + \angleOEC

\Rightarrow \angle OEB=90 \degree-\frac{z}{2}+z                  (from  3)

\Rightarrow \angle OEB=90 \degree+\frac{z}{2}.................................................................6

From 4,5,6 ;we get

            \angleBDE = \angleBED = \angleOEB - \angleOED

\Rightarrow \angle BDE=\angle BED=90 \degree+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}

\Rightarrow \angle BDE+\angle BED=y+z..................................................7

In \triangleBDE , 

         \angleDBE + \angleBDE + \angleBED = 180 \degree

\Rightarrow \angle DBE +y+z=180 \degree

\Rightarrow \angle DBE =180 \degree-(y+z)

\Rightarrow \angle ABC =180 \degree-(y+z)...................................................8

Here,  from equation 1,

         \frac{x-y}{2}=\frac{360 \degree-y-2x-y}{2}

\Rightarrow \frac{x-y}{2}=\frac{360 \degree-2y-2x}{2}

\Rightarrow \frac{x-y}{2}=180 \degree-y-x...................................9

From 8 and 9,we have 

      \angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer:

Given : ABCD is rhombus.

To prove : the circle drawn with AB  as diameter, passes through the  point O.

Proof : 

        

         ABCD is rhombus.

 Thus, \angle AOC = 90 \degree             (diagonals of a rhombus bisect each other at 90 \degree)

So, a circle drawn AB as diameter will pass through point O.

Thus,  the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Answer:

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove :   AE = AD 

Proof : 

              

          \angleADC = \angle3  , \angleABC = \angle4, \angleADE = \angle1  and \angleAED = \angle2

           \angle 3+\angle 1=180 \degree.................1(linear pair)

         \angle 2+\angle 4=180 \degree....................2(sum of opposite angles of cyclic quadrilateral)

         \angle3 = \angle4      (oppsoite angles of parallelogram )

 From 1 and 2,

           \angle3+\angle1 = \angle2 + \angle

From 3,    \angle1 = \angle

From 4,    \triangleAQB,     \angle1 = \angle2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA. 

Proof :

               

In \triangle ABD and \triangleCDO,

          AO = OC     (Given )

         \angleAOB = \angleCOD  (Vertically opposite angles )

          BO = DO        (Given )

So,  \triangle ABD\cong \triangleCDO     (By SAS)

       \angleBAO = \angleDCO    (CPCT)

\angleBAO and \angleDCO are alternate angle and are equal .

So,     AB || DC ..............1

   Also  AD || BC  ...............2

From 1 and 2,

            \angle A+\angle C=180 \degree......................3(sum of opposite angles)

           \angleA = \angleC    ................................4(Opposite angles of the parallelogram )

From 3 and 4,

    \angle A+\angle A=180 \degree

\Rightarrow 2\angle A=180 \degree

 \Rightarrow \angle A=90 \degree

BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Answer:

Given:  AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA. 

Proof :

                  

       ABCD is a parallelogram. (proved in (i))

       \angle A=90 \degree                    (proved in (i))

A parallelogram with one angle 90 \degree, is a rectangle )

Thus, ABCD is rectangle.

Answer:

Given :   Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove :  the angles of the triangle DEF are    \small 90^{\circ}-\frac{1}{2}C,    \small 90^{\circ}-\frac{1}{2}B and \small 90^{\circ}-\frac{1}{2}A

Proof : 

          

\angle1 and \angle3 are angles in same segment.therefore,

          \angle1 = \angle3 ................1(angles in same segment are equal )

and    \angle2 = \angle4 ..................2

Adding 1 and 2,we have 

         \angle1+\angle2=\angle3+\angle4

\Rightarrow \angle D=\frac{1}{2}\angle B+\frac{1}{2}\angle C,

\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)

\Rightarrow \angle D=\frac{1}{2}(180 \degree+\angle C)

and  \Rightarrow \angle D=\frac{1}{2}(180 \degree-\angle A)

        \Rightarrow \angle D=90 \degree-\frac{1}{2}\angle A

Similarly,  \Rightarrow \angle E=90 \degree-\frac{1}{2}\angle B     and    \angle F=90 \degree-\frac{1}{2}\angle C

Q9 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that \small BP=BQ.

Answer:

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove :  BP = BQ 

Proof :

         

AB is a common chord in both congruent circles.

  \therefore \angle APB = \angle AQB

In \triangle BPQ,

   \angle APB = \angle AQB

    \therefore BQ = BP      (Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of  \small \angle A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Given :In any triangle ABC, if the angle bisector of  \small \angle A and perpendicular bisector of BC  intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

           

Let \angleABD = \angle1 ,  \angleADC = \angle2 , \angleDCB =\angle3 , \angleCBD = \angle4

\angle1 and \angle3 lies in same segment.So,

        \angle1 = \angle3    ..........................1(angles in same segment)

similarly, \angle2 = \angle4  ......................2

also,       \angle1=\angle2 ..............3(given)

From 1,2,3 , we get

     \angle3 = \angle4

Hence,  BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 10 Circles

  • Learn and memorize some theorems related to circles

  • Learn the application of those theorems in the problems

  • Start applying the theorems and concepts on the practice exercises.

  • During the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 10 Circles.

  • After doing all the above-written things, you can practice more using the previous year questions papers.

Keep working hard & happy learning!

 

Recently Asked Questions

 

Related Articles

Exams
Articles
Questions